The gravitational force between two stones of mass 1 kg each separated by a distance of

1 metre in vacuum is

1. Zero                                                 

2. 6.675$\times {10}^{‐5}newton$

3. $6.675\times {10}^{‐11}newton$                 

4. $6.675\times {10}^{‐8}newton$

The escape velocity for the Earth is taken $v_d$. Then, the escape velocity for a planet whose radius is four times and the density is nine times that of the earth, is:

Two particles of equal mass go round a circle of radius R under the action of their mutual

gravitational attraction. The speed of each particle is

1. $\nu =\frac{1}{2R}\sqrt{\frac{1}{Gm}}$                             

2. $\nu =\sqrt{\frac{Gm}{2R}}$

3. $\nu =\frac{1}{2}\sqrt{\frac{Gm}{R}}$                               

4. $\nu =\sqrt{\frac{4Gm}{R}}$   

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R

from the surface of the earth is (g = acceleration due to gravity at the surface of the

earth)

1. $\frac{g}{9}$                       

2. $\frac{g}{3}$

3. $\frac{g}{4}$                       

4.  g

 If V, R, and g denote respectively the escape velocity from the surface of the earth, the

radius of the earth, and acceleration due to gravity, then the correct equation is:

1. $V=\sqrt{gR}$                                 

2. V=$\sqrt{\frac{4}{3}g{R}^3}$

3. V = R$\sqrt{g}$                                   

4. V=$\sqrt{2gR}$

The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$, is:

1. R                              

2. $\frac{3R}{4}$

3. $\frac{R}{2}$                             

4. $\frac{R}{4}$

The value of $g$ at a particular point is $9.8~\text{m/s}^2$. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass then value of $g$ at the same point will now become: (assuming that the distance of the point from the centre of the earth does not shrink)

The earth (mass = $6\times {10}^{24}kg$ ) revolves round the sun with angular velocity

$2\times {10}^{‐7}rad/s$ in a circular orbit of radius $1.5\times {10}^{8}km$ . The force exerted by the sun on

the earth in Newtons, is

1. $18\times {10}^{25}$                               

2. Zero

3. $27\times {10}^{39}$                               

4. $36\times {10}^{21}$